1443-501 Spring 2002
Lecture #19
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
The Pendulum
Physical Pendulum
Simple Harmonic and Uniform Circular Motions
Damped Oscillation
Review Examples Ch. 10-13
No Homework Assignment today!!!!!
2nd term exam on Wednesday, Apr. 10. Will cover chapters 10 -13.
The Pendulum
A simple pendulum also performs periodic motion.
The net force exerted on the bob is
 Fr  T  mg cos A  0
L

m
s
d 2s
 Ft  mg sin  A  ma  m dt 2
T
mg
Since the arc length, s, is s  L A
d 2s
d 2
d 2
g

L


g
sin



sin 
results
2
2
2
dt
dt
dt
L
Again became a second degree differential equation,
satisfying conditions for simple harmonic motion
d 2
g
g
2









If  is very small, sin~
giving
angular
frequency
2
dt
The period for this motion is T 
Apr. 8, 2002
L
2

 2
L
L
g
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
The period only depends on the
length of the string and the
gravitational acceleration
2
Example 13.5
Christian Huygens (1629-1695), the greatest clock maker in history, suggested that an
international unit of length could be defined as the length of a simple pendulum having a
period of exactly 1s. How much shorter would out length unit be had this suggestion
been followed?
Since the period of a simple
2
T
 2
pendulum motion is

L
g
The length of the pendulum
in terms of T is
T 2g
L
4 2
Thus the length of the
pendulum when T=1s is
T 2 g 1 9.8
L

 0.248m
2
2
4
4
Therefore the difference in
length with respect to the
current definition of 1m is
L  1  L  1  0.248  0.752m
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
3
Physical Pendulum
Physical pendulum is an object that oscillates about a fixed
axis which does not go through the object’s center of mass.
O

Consider a rigid body pivoted at a point O that is a distance d from the CM.
d
dsin
CM
mg
The magnitude of the net torque provided by the gravity is
  mgd sin 
Then
d 2
  I  I dt 2  mgd sin 
mgd
Therefore, one can rewrite d 2
 mgd 



sin










2
dt
I
 I 
Thus, the angular frequency  is

And the period for this motion is T 
Apr. 8, 2002
2

mgd
I
 2
I
mgd
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
By measuring the period of
physical pendulum, one can
measure moment of inertia.
Does this work for
simple pendulum?
4
Example 13.6
A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical
plane. Find the period of oscillation if the amplitude of the motion is small.
O
Pivot
L
CM
1
Moment of inertia of a uniform rod,
I  ML2
rotating about the axis at one end is
3
The distance d from the pivot to the CM is L/2,
therefore the period of this physical pendulum is
Mg
T
2

 2
I
2ML2
 2
 2
Mgd
3MgL
2L
3g
Calculate the period of a meter stick that is pivot about one end and is oscillating in a
vertical plane.
Since L=1m,
the period is
Apr. 8, 2002
T  2
2L
 2
3g
2
 1.64 s
3  9.8
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
1
So the
f   0.61s 1
frequency is
T
5
Torsional Pendulum
When a rigid body is suspended by a wire to a fixed support at the top and the
body is twisted through some small angle , the twisted wire can exert a restoring
torque on the body that is proportional to the angular displacement.
The torque acting on the body due to the wire is
O
P
  
max
Applying the Newton’s second
law of rotational motion
 is the torsion
constant of the wire
d 2
  I  I dt 2  
d 2
 

Then, again the equation becomes









2
dt
I
Thus, the angular frequency  is
And the period for this motion is
Apr. 8, 2002

T

I
2

 2
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
I

This result works as
long as the elastic limit
of the wire is not
exceeded
6
Simple Harmonic and Uniform Circular Motions
Uniform circular motion can be understood as a
superposition of two simple harmonic motions in x and y axis.
y
y
A
O
P
y

x
P
y

A
y

A a

O
Q
A

O x Q
O vx Q
x
t=t
t=0
=t+
When the particle rotates at a uniform angular
speed , x and y coordinate position become
Since the linear velocity in a uniform circular
motion is A, the velocity components are
Since the radial acceleration in a uniform circular
motion is v2/A=A, the components are
Apr. 8, 2002
v
P
x
ax P
x  A cos   A cost   
y  A sin   A sin t   
vx  v sin    A sin t   
v y  v cos   A cost   
a x   a cos    A 2 cost   
a y   a sin    A 2 sin t   
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
7
x
Example 13.7
A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular
speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to
the right. A) Determine the x coordinate as a function of time.
Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the
angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is
x  A cos  3.00mcos8.00t   
Since x=2.00, when t=0
 2.00 

2.00  3.00m  cos  ;   cos 1 
  48.2
 3.00 

However, since the particle was

x

3
.
00
m
cos
8
.
00
t

48
.
2
moving to the right =-48.2o,
Find the x components of the particle’s velocity and acceleration at any time t.

Using the
displcement
Likewise,
from velocity
Apr. 8, 2002




 

dx
vx   3.00  8.00sin 8.00t  48.2   24.0m / s sin 8.00t  48.2
dt
dv
a x    24.0  8.00 cos8.00t  48.2   192m / s 2 cos 8.00t  48.2
dt

1443-501 Spring 2002
Dr. J. Yu, Lecture #19
8
Damped Oscillation
More realistic oscillation where an oscillating object loses its mechanical
energy in time by a retarding force such as friction or air resistance.
F
Let’s consider a system whose retarding force
is air resistance R=-bv (b is called damping
coefficient) and restoration force is -kx
The solution for the above 2nd order
differential equation is
The angular frequency 
for this motion is

x
 kx  bv  max
dx
d 2x
 kx  b
m 2
dt
dt
x  Ae
k  b 


m  2m 

b
t
2m
cost   
2
This equation of motion tells us that when the retarding force is much smaller than restoration
force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops.
We express the
 b 
     

angular frequency as
2
m


Apr. 8, 2002
2
Where the natural
0 
frequency 0
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
k
m
9
More on Damped Oscillation
The motion is called Underdamped when the magnitude of
the maximum retarding force Rmax = bvmax <kA
How do you think the damping motion would change as
retarding force changes?
As the retarding force become larger, the amplitude reduces
more rapidly, eventually stopping its equilibrium position
 0
Under what condition this system
b
 
does not oscillate?
 bvmax  kA
The system is Critically damped
What do you think happen?
2m
b  2m   2 mk
Once released from non-equilibrium position, the object
would return to its equilibrium position and stops.
If the retarding force is larger
Rmax  bvmax  kA The system is Overdamped
than restoration force
Once released from non-equilibrium position, the object would return
Spring
2002stops, but a lot slower than before10
Apr. 8, 2002
to its equilibrium1443-501
position
and
Dr. J. Yu, Lecture #19
Rotational Kinematics
The first type of motion we have learned in linear kinematics was
under a constant acceleration. We will learn about the rotational
motion under constant acceleration, because these are the simplest
motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant
angular acceleration:
 f   i  t
Angular displacement under
constant angular acceleration:
1 2
 f   i   i t  t
2
One can also obtain
    2  f   i 
Apr. 8, 2002
2
f
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
2
i
11
Example 10.1
A wheel rotates with a constant angular acceleration pf 3.50 rad/s2. If
the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what
angle does the wheel rotate in 2.00s?
Using the angular displacement
formula in the previous slide, one gets
What is the angular speed at t=2.00s?
Using the angular speed and
acceleration relationship
Find the angle through which the wheel
rotates between t=2.00 s and t=3.00 s.
Apr. 8, 2002
1 2
t
2
1
2
 2.00  2.00  3.50  2.00 
2
11.0
 11.0rad 
rev.  1.75rev.
2
 f   i  t 
 f   i  t
 2.00  3.50  2.00
 9.00rad / s
 2  11.0rad
1
2
3.50  3.00   21.8rad
2
10.8
      2  10.8rad 
rev.  1.72rev.
2
 3  2.00  3.00 
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
12
Rotational Energy
y
vi
mi
ri

What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
Kinetic energy of a masslet, mi,
1
1
K i  mi vi2  mi ri 2 
2
2
moving at a tangential speed, vi, is
x
O
Since a rigid body is a collection of masslets,
the total kinetic energy of the rigid object is
By defining a new quantity called,
Moment of Inertia, I, as
K R   Ki 
I   mi ri2
i
i
The above expression
is simplified as
What are the dimension and unit of Moment of Inertia?
What do you think the
moment of inertia is?
1
1
2 
2 
m
r


m
r



i i
i i 
2 i
2 i

kg m 2
ML 
1
K R  I 
2
2
Measure of resistance of an object to
changes in its rotational motion.
What similarity do you see between
Mass and speed in linear kinetic energy are
rotational and linear kinetic energies? replaced by moment of inertia and angular speed.
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
13
Example 10.4
In a system consists of four small spheres as shown in the figure, assuming the radii are
negligible and the rods connecting the particles are massless, compute the moment of
inertia and the rotational kinetic energy when the system rotates about the y-axis at .
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
b
M
x
I   mi ri 2  Ml 2  Ml 2  m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
Why are some 0s?
m
and the radii of the spheres are negligible.
1 2 1
K R  I  2 Ml 2  2  Ml 2 2
Thus, the rotational kinetic energy is
2
2


Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.
I   mi ri 2  Ml 2  Ml 2  mb2  mb2  2Ml 2  mb2 
i
Apr. 8, 2002




1
1
K R  I 2  2Ml 2  2mb2  2  Ml 2  mb2  2
2
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
14
Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume
the object consists of small volume elements with mass, mi.
ri 2 mi   r 2 dm

The moment of inertia for the large rigid object isI  lim
m 0
i
i
It is sometimes easier to compute moments of inertia in terms of volume of the elements
rather than their mass
dm
Using the volume density, r, replace
I  rr 2 dV
r  ; dm  rdV The moments of
dm in the above equation with dV.
inertia becomes
dV

Example 10.5: Find the moment of inertia of a uniform hoop of mass M and radius R
about an axis perpendicular to the plane of the hoop and passing through its center.
y
O
Apr. 8, 2002
The moment
of inertia is
dm
R
x
I   r 2 dm  R 2  dm  MR 2
What do you notice
from this result?
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
15
Example 10.6
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis perpendicular to the rod and passing through its center of mass.

The line density of the rod is
y
dm  dx 
so the masslet is
dx
x
x
L
The moment
of inertia is
Will this be the same as the above.
Why or why not?
Apr. 8, 2002
M
dx
L
L/2
x2M
M 1 3 
2
I   r dm  
dx 
x 

L / 2
L
L  3  L / 2
L/2
M  L   L 

     
3L  2   2 
3
What is the moment of inertia
when the rotational axis is at
one end of the rod.
M
L
I   r dm  
2
L
0


3
 M  L3  ML2
  

3
L

 4  12
L
x2M
M 1 3 
dx 
x
L
L  3  0
 
M
M 3
ML2
3
L   0 

L 
3L
3L
3
Since the moment of inertia is resistance to motion, it makes perfect sense
for it to be harder to move when it is rotating about the axis at one end.
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
16
Parallel Axis Theorem
Moments of inertia for highly symmetric object is relatively easy if the
rotational axis is the same as the axis of symmetry. However if the axis of
rotation does not coincide with axis of symmetry, the calculation can still be
2
done in simple manner using parallel-axis theorem. I  I CM  MD
y
Moment of inertia is defined I   r 2 dm   x 2  y 2 dm (1)
(x,y)
Since x and y are
x  xCM  x' ;
y  yCM  y'
r
yCM
y
y’
One can substitute x and y in Eq. 1 to obtain
CM
(xCM,yCM)
D
xCM
x’
x
What does this
theorem
tell you?
Apr. 8, 2002


I   xCM  x'   yCM  y ' dm

2

2


2
2
2
2
 xCM
 yCM
dm

2
x
x
'
dm

2
y
y
'
dm

x
'

y
'
dm
CM
CM




 x' dm  0
x
 y ' dm  0
2
I

I

MD
Therefore, the parallel-axis theorem
CM
Since the x’ and y’ are the
distance from CM, by definition
Moment of inertia of any object about any arbitrary axis are the same as
the sum of moment
inertia
of the
1443-501ofSpring
2002for a rotation about the CM and that 17
J. Yu, Lecture
CM about theDr.
rotation
axis. #19
Example 10.8
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis that goes through one end of the rod, using parallel-axis theorem.

The line density of the rod is
y
dm  dx 
so the masslet is
CM
dx
M
L
M
dx
L
L/2
x2M
M 1 3 
2
  r dm  
dx 
x 

L / 2
L
L  3  L / 2
x The moment of I CM
inertia about
3
3
the CM
M  L   L  
L/2
x
L
M  L3  ML2
 

       
3L  2   2   3L  4  12
2
Using the parallel axis theorem
I  I CM
ML2  L 
ML2 ML2 ML2
2
D M 
  M 


12  2 
12
4
3
The result is the same as using the definition of moment of inertia.
Parallel-axis theorem is useful to compute moment of inertia of a rotation of a
rigid object with complicated shape about an arbitrary axis
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
18
Torque
Torque is the tendency of a force to rotate an object about some axis.
Torque, , is a vector quantity.
F

r
P
Line of
Action
d2
d
Moment
arm
F2
Consider an object pivoting about the point P
by the force F being exerted at a distance r.
The line that extends out of the tail of the force
vector is called the line of action.
The perpendicular distance from the pivoting point
P to the line of action is called Moment arm.
Magnitude of torque is defined as the product of the force
exerted on the object to rotate it and the moment arm.
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force vectorially. The convention for sign of the torque is positive if
rotation is in counter-clockwise and negative if clockwise.
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
  rF sin   Fd
  
1
 2
 Fd  F2 d 2
19
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating in a circle.
What forces do you see in this motion?
m
The tangential force Ft and radial force Fr
Ft  mat  mr
The tangential force Ft is
The torque due to tangential force Ft is
What do you see from the above relationship?
What does this mean?
  Ft r  mat r  mr 2
  I
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2nd law of motion in rotation.
How about a rigid object?
The external tangential force dFt is dFt  dmat  dmr
dFt
The torque due to tangential force Ft is d  dFt r  r 2 dm 
dm
2



r
dm  I
The total torque is


r
Contribution from radial force is 0, because its
What is the contribution due
line of action passes through the pivoting
O
to
radial
force
and
why?
1443-501 Spring 2002 point, making the moment arm 0.
Apr. 8, 2002
20
Dr. J. Yu, Lecture #19
Example 10.10
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is
free to rotate about the pivot in the vertical plane. The rod is released from rest in the
horizontal position what is the initial angular acceleration of the rod and the initial linear
acceleration of its right end?
The only force generating torque is the gravitational force Mg
L/2
  Fd  F
Mg
L
L
 Mg  I
2
2
L
Since the moment of inertia of the rod I  r 2 dm 
0
0
when it rotates about one end is
L
We obtain

MgL MgL 3g


2
2ML 2 L
2I
3
Apr. 8, 2002
L
3
2


M
x
ML


x 2dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
21
Work, Power, and Energy in Rotation
F

d r
ds
Let’s consider a motion of a rigid body with a single external
force F exerted on the point P, moving the object by ds.
The work done by the force F as the object rotates
through infinitesimal distance ds=rd in a time dt is
dW  F  d s  F sin  rd
O
What is Fsin?
What is the work done by
radial component Fcos?
Since the magnitude of torque is rFsin,
The tangential component of force F.
Zero, because it is perpendicular to the
displacement.
dW  d
P
The rate of work, or power becomes
dW d

 
dt
dt
How was the power
defined in linear motion?
 d 
 d  d 

I




dt
d

dt





The rotational work done by an external force
equals the change in rotational energy.
  I  I 
The work put in by the external force then
dW   d  Id
Apr. 8, 2002

W  
1443-501 Spring 2002
Dr. J. Yu, Lecture #19

f
 d   Id 

1 2 1 2
I i  I f
2
2 22
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Similar Quantity
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Apr. 8, 2002
Linear
Mass
Rotational
Moment of Inertia
M
Distance
I   r 2 dm
L
Angle  (Radian)
v 
dr
dt
 
d
dt
a 
dv
dt
 
d
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Torque
Work
  I
W 
P  F v
P  
p  mv
L  I
K 
1
mv 2
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
Rotational
f

d 
i
KR 
1
I 2
2
23
Rolling Motion of a Rigid Body
What is a rolling motion?
A more generalized case of a motion where the
rotational axis moves together with the object
A rotational motion about the moving axis
To simplify the discussion, let’s
make a few assumptions
1.
2.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
The object rolls on a flat surface
Let’s consider a cylinder rolling without slipping on a flat surface
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
s  R
R  s
s=R
Apr. 8, 2002
Thus the linear
speed of the CM is
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
vCM 
ds
d
 R  R
dt
dt
Condition for “Pure Rolling”
24
Total Kinetic Energy of a Rolling Body
What do you think the total kinetic
energy of the rolling cylinder is?
P’
CM
Since it is a rotational motion about the point
P, we can writ the total kinetic energy
1
K  I P 2
2
2vCM
vCM
Using the parallel axis theorem, we can rewrite
P
K
Since vCM=R, the above
relationship can be rewritten as
What does this equation mean?
1
1
1
I P 2  I CM  2  MR 2 2
2
2
2
1
1
2
2
K  I CM   MvCM
2
2
Rotational kinetic
energy about the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Apr. 8, 2002
Where, IP, is the moment of
inertia about the point P.
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
25
Example 11.1
For solid sphere as shown in the figure, calculate the linear speed of the CM at the
bottom of the hill and the magnitude of linear acceleration of the CM.
What must we know first?
R

h
I CM   r 2 dm 

vCM
vCM 
Since h=xsin,
one obtains
2
vCM

The linear acceleration
of the CM is
Apr. 8, 2002
2
MR 2
5
Thus using the formula in the previous slide
2 gh

2
1  I CM / MR
10
gx sin 
7
aCM
The moment of inertia the
sphere with respect to the CM!!
Using kinematic
relationship
2
vCM
5

 g sin 
2x
7
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
2 gh

1 2 / 5
10
gh
7
2
vCM
 2aCM x
What do you see?
Linear acceleration of a sphere does
not depend on anything but g and .
26
Torque and Vector Product
z
O
r
rxF
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
p
The disk will start rotating counter clockwise about the Z axis
y
The magnitude of torque given to the disk by the force F is

  Fr sin 
F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
What is the direction?
The direction of the torque follows the right-hand rule!!
The above quantity is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Apr. 8, 2002
  rF
C  A B
C  A  B  A B sin 
What is another vector operation we’ve learned?
Scalar product
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
C  A  B  A B cos
Result? A scalar
27
Angular Momentum of a Particle
If you grab onto a pole while running, your body will rotate about the pole, gaining
angular momentum. We’ve used linear momentum to solve physical problems
with linear motions, angular momentum will do the same for rotational motions.
z
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
L=rxp
The instantaneous angular momentum
L  r p
L of this particle relative to origin O is
O
y
m
r
What is the unit and dimension of angular momentum? kg m2 / s 2
 p
Note that L depends on origin O. Why? Because r changes
x
What else do you learn? The direction of L is +z
Since p is mv, the magnitude of L becomes L  mvr sin 
What do you learn from this?
The point O has
to be inertial.
Apr. 8, 2002
If the direction of linear velocity points to the origin of
rotation, the particle does not have any angular momentum.
If the linear velocity is perpendicular to position vector, the
particle
moves
the same way as a point on a28rim.
1443-501
Springexactly
2002
Dr. J. Yu, Lecture #19
Angular Momentum and Torque
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
F 
dp
dt
Total external forces exerting on a particle is the same as the change of its linear momentum.
The same analogy works in rotational motion between torque and angular momentum.
Net torque acting on a particle is   r   F  r  d p
dt
z
dL
d r p
dr
dp
dp


 pr
 0r
L=rxp
dt
dt
dt
dt
dt

O
r
x
y
m


Why does this work?
p
Thus the torque-angular
momentum relationship
Because v is parallel to
the linear momentum
 
dL
dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
29
Example 11.4
A particle of mass m is moving in the xy plane in a circular path of radius r and linear
velocity v about the origin O. Find the magnitude and direction of angular momentum with
respect to O.
Using the definition of angular momentum
y
v
L  r  p  r  mv  mr  v
r
O
x
Since both the vectors, r and v, are on x-y plane and
using right-hand rule, the direction of the angular
momentum vector is +z (coming out of the screen)
The magnitude of the angular momentum is L  mr  v  mrv sin   mrv sin 90  mrv
So the angular momentum vector can be expressed as
L  mrvk
Find the angular momentum in terms of angular velocity .
Using the relationship between linear and angular speed
L  mrvk  mr 2 k  mr 2   I 
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
30
Example 11.6
A rigid rod of mass M and length l pivoted without friction at its center. Two particles of mass
m1 and m2 are connected to its ends. The combination rotates in a vertical plane with an
angular speed of . Find an expression for the magnitude of the angular momentum.
y
m2
l

O
m1
m1 g
m2 g
x
Find an expression for the magnitude of the angular acceleration of the
system when the rod makes an angle  with the horizon.
If m1 = m2, no angular
momentum because net
torque is 0.
If //, at equilibrium
so no angular momentum.
Apr. 8, 2002
The moment of inertia of this system is
1
1
1
2
2
I  I rod  I m1  I m2  Ml  m1l  m2l 2
12
4
4
2
l2  1

  M  m1  m2  L  I  l  1 M  m1  m2 
4 3

4 3

First compute net
external torque
   m1 g
l
l
cos  ;  2  m2 g cos 
2
2
 ext      2 
gl cos m1  m2 
2
1
m1  m1 gl cos 
 ext
2m1  m1  cos 

Thus 
2

 2

g /l
l 1
I
1


becomes 1443-501 Spring 2002  M  m1  m2   M  m1  m
312 
4 3
 3
Dr. J. Yu, Lecture #19
Example 11.8
A start rotates with a period of 30days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses
into a neutron start of radius 3.0km. Determine the period of rotation of the neutron star.
The period will be significantly shorter,
because its radius got smaller.
1. There is no torque acting on it
Let’s make some assumptions:
2. The shape remains spherical
3. Its mass remains constant
Li  L f
Using angular momentum
conservation
I i  I f  f
What is your guess about the answer?
The angular speed of the star with the period T is
Thus
2

T
I i
mri 2 2
f 

If
mrf2 Ti
Tf 
Apr. 8, 2002
2
f
 r f2
 2
r
 i
2

3
.
0


6
Ti  

30
days

2
.
7

10
days  0.23s

4

 1.0  10 

1443-501 Spring 2002
Dr. J. Yu, Lecture #19
32
Conditions for Equilibrium
What do you think does the term “An object is at its equilibrium” mean?
The object is either at rest (Static Equilibrium) or its center of mass
is moving with a constant velocity (Dynamic Equilibrium).
When do you think an object is at its equilibrium?
Translational Equilibrium: Equilibrium in linear motion
Is this it?
The above condition is sufficient for a point-like particle to be at its static
equilibrium. However for object with size this is not sufficient. One more
condition is needed. What is it?
Let’s consider two forces equal magnitude but opposite direction acting
on a rigid object as shown in the figure. What do you think will happen?
F
d
d
F 0
CM
-F
Apr. 8, 2002

The object will rotate about the CM. The net torque
 0
acting on the object about any axis must be 0.
For an object to be at its static equilibrium, the object should not
have linear or angular speed. v
0  0
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
CM
33
More on Conditions for Equilibrium
To simplify the problems, we will only deal with forces acting on x-y plane, giving torque
only along z-axis. What do you think the conditions for equilibrium be in this case?
The six possible equations from the two vector equations turns to three equations.
F 0 F
F
x
0
y
0
  0 
0
z
What happens if there are many forces exerting on the object?
If an object is at its translational static equilibrium, and if the
net torque acting on the object is 0 about one axis, the net
torque must be 0 about any arbitrary axis.
r’
r5 O O’
Net Force exerting on the object  F  F 1  F 2  F 3      0

Net torque about O
O
 r 1  F 1  r 2  F 2  r 3  F 3       r i  F i 0
Position of force Fi about O’
Net torque about O’
Apr. 8, 2002

O'
 
'
i
r  r i  r'


 r '1F 1  r '2  F 2      r 1  r '  F 1  r 2  r '  F 2       r i  F i  r ' F i
1443-501
  rSpring
 F 2002
 r '0 
Dr.
J. Yu, Lecture #19
O'
i
i
O
0
34
Example 12.1
A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N,
respectively. If the support (or fulcrum) is under the center of gravity of the board and the
father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board
by the support?
1m
F
MFg
x
n
MBg
D
MFg
Since there is no linear motion, this system
is in its translational equilibrium
F
F
Therefore the magnitude of the normal force
x
0
y
 MBg  MF g  MDg  n  0
n  40.0  800  350  1190N
Determine where the child should sit to balance the system.
The net torque about the fulcrum   M g  0  M g 1.00  M g  x  0
B
F
D
by the three forces are
Therefore to balance the system x  M F g 1.00m  800 1.00m  2.29m
the daughter must sit
MDg
350
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
35
Example 12.1 Continued
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg
x
n
x/2
MBg
D
MFg
The net torque about the axis of
rotation by all the forces are
  M B g  x / 2  M F g  1.00  x / 2  n  x / 2  M D g  x / 2  0
Since the normal force is
n  MBg  MF g  MDg
The net torque can   M B g  x / 2  M F g  1.00  x / 2 
be rewritten
 M B g  M F g  M D g   x / 2  M D g  x / 2
 M F g 1.00  M D g  x  0
Therefore
Apr. 8, 2002
x
MFg
800
1.00m 
1.00m  2.29m
MDg
350
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
36
Example 12.3
A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by
a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the
horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as
the magnitude and direction of the force exerted by the wall on the beam.
R

53.0o
53.0o
FBD
Rsin
Rcos
8m
From the rotational equilibrium
Using the
translational
equilibrium
Apr. 8, 2002
First the translational equilibrium,
using components
N
6N
2m
T
F
F
Tsin53
Tcos53
 R cos  T cos 53.0  0
y
R sin   T sin 53.0  600 N  200 N  0
 T sin 53.0  8.00  600 N  2.00  200 N  4.00m  0

T  313N
R cos  T cos 53.0
And the magnitude of R is
R sin   T sin 53.0  600 N  200 N
 800  313  sin 53.0 


  tan 

71
.
7


 313 cos 53.0

1
x
T cos 53.0 313  cos 53.0
R

 582 N
cos
cos 71.1
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
37
Example 12.4
A uniform ladder of length l and weight mg=50 N rests against a smooth, vertical wall. If
the coefficient of static friction between the ladder and the ground is ms=0.40, find the
minimum angle min at which the ladder does not slip.
First the translational equilibrium,
using components
P
l
FBD

mg
n
O
F
F
f
Thus, the normal force is
x
f  P  0
y
 mg  n  0
n  mg  50 N
The maximum static friction force
just before slipping is, therefore,
f smax  m s n  0.4  50 N  20 N  P
From the rotational equilibrium

 min
Apr. 8, 2002
l
  mg cos min  Pl sin  min  0
2
 mg 
1  50 N 

 tan 1 
  tan 
  51
 2P 
 40 N 
O
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
38
Example 12.7
A solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The
sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The
volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is
submerged?
Since bulk modulus is
P
B
V
Vi
The amount of volume change is
V  
PVi
B
From table 12.1, bulk modulus of brass is 6.1x1010 N/m2
The pressure change P is
P  Pf  Pi  2.0 107 1.0 105  2.0 107
Therefore the resulting
2.0 107  0.5
4
3

V

V

V




1
.
6

10
m
f
i
volume change V is
6.11010
The volume has decreased.
Apr. 8, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #19
39